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<title>Problem 2091. -- [Poi2010]The Minima Game -- 衡阳八中OJ离线版-2012-02-29</title><center><h2>2091: [Poi2010]The Minima Game</h2><span class=green>Time Limit: </span>10 Sec&nbsp;&nbsp;<span class=green>Memory Limit: </span>259 MB<br><span class=green>Submit: </span>156&nbsp;&nbsp;<span class=green>Solved: </span>102<br>[<a href='submitpage.php?id=2091'>Submit</a>][<a href='problemstatus.php?id=2091'>Status</a>][<a href='bbs.php?id=2091'>Discuss</a>]</center><h2>Description</h2><div class=content><p>给出N个正整数，AB两个人轮流取数，A先取。每次可以取任意多个数，直到N个数都被取走。<br />
每次获得的得分为取的数中的最小值，A和B的策略都是尽可能使得自己的得分减去对手的得分更大。<br />
在这样的情况下，最终A的得分减去B的得分为多少。</p>
<p></p></div><h2>Input</h2><div class=content><p>第一行一个正整数N (N &lt;= 1,000,000)，第二行N个正整数（不超过10^9）。</p>
<p></p></div><h2>Output</h2><div class=content><p><br />
一个正整数，表示最终A与B的分差。</p>
<p></p></div><h2>Sample Input</h2>
			<div class=content><span class=sampledata>3<br />
1 3 1<br />
<br />
</span></div><h2>Sample Output</h2>
			<div class=content><span class=sampledata>2<br />
<br />
</span></div><h2>HINT</h2>
			<div class=content><p><p>第一次A取走3，第二次B取走两个1，最终分差为2。<br /><br />
</p></p></div><h2>Source</h2>
			<div class=content><p><a href='problemset.html?search=鸣谢 JZP'>鸣谢 JZP</a></p></div><center>[<a href='submitpage.php?id=2091'>Submit</a>][<a href='problemstatus.php?id=2091'>Status</a>][<a href='bbs.php?id=2091'>Discuss</a>]</center>﻿<br>

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